Question

What is the time taken for a mango to fall 20m from it tree (g=10m/s)​

Answer 1

`\boxed {\boxed {\sf 2 \ seconds}}`

Step-by-step explanation:

We are asked to find the time it takes for a mango to fall 20 meters.

We know the distance, acceleration, and initial velocity, so we will use the following kinematic equation:

`d= v_i t + (1)/(2) at^2`

The mango is dropped from rest, so the initial velocity is 0 meters per second. It falls a distance of 20 meters. The acceleration due to gravity is 10 meters per second squared.

• 
  `v_i`= 0 m/s
• d= 20 m
• a= 10 m/s²

Substitute the values into the equation.

`20 \ m = (0 \ m/s)(t) + (1)/(2) (10 \ m/s^2)(t^2)`

`20 \ m = (1)/(2) (10 \ m/s^2)(t^2)`

`20 \ m = (5 \ m/s^2)(t^2)`

We are solving for time, so we must isolate the variable t. It is being multiplied by 5 meters per second squared. The inverse operation of multiplication is division, so divide both sides by 5 m/s².

`\frac {20 \ m}{5 \ m/s^2}= ((5 \ m/s^2)(t^2))/(5 \ m/s^2)`

`\frac {20 \ m}{5 \ m/s^2}=t^2`

`4 \ s^2=t^2`

The variable t is being squared. Take the square root of both sides.

`\sqrt { 4 \ s^2 }= √(t^2)`

`2 \ s=t`

It takes 2 seconds for a mango to fall 20 meters.

Answer 2

Hi there!

We can use the kinematic equation:

`d = v_0t + (1)/(2)at^2`

d = displacement (20 m)

v0 = initial velocity (dropped from rest, so 0 m/s)

t = time (s)

a = acceleration due to gravity (10 m/s²)

Rearrange the equation to solve for time:

`d = 0 + (1)/(2)at^2\\\\2d = at^2\\\\t^2 = (2d)/(a)\\\\t = \sqrt{(2d)/(a)}`

Solve using the given values:

`t = \sqrt{(2(20))/(10)} = √(4) = \boxed{2 sec}`